Question

How do you find the equation of the tangent line to the graph of `f(x)=sqrtx` at point (1,1)?

Answer 1

`y=1/2x+1/2`

Explanation:

The equation of the tangent in `color(blue)"point-slope form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1)" is a point on the tangent"`

`color(orange)"Reminder: " m_("tangent")=dy/dx" at x=a"`

`f(x)=sqrtx=x^(1/2)`

differentiate using the `color(blue)"power rule"`

`rArrf'(x)=1/2x^(-1/2)=1/(2sqrtx)`

`rArrf'(1)=1/2=m_("tangent")`

`"Using " m=1/2" and " (x_1,y_1)=(1,1)" then"`

`rArry-1=1/2(x-1)`

`rArry=1/2x-1/2+1`

`rArry=1/2x+1/2" is equation of tangent line"`